AP Chemistry 7.2: Equilibrium Constants, Expressions, and Reaction Quotients

Once a reaction reaches equilibrium, the relative amounts of products and reactants can be described mathematically using an equilibrium constant, represented by K.

• For the general reaction:
• aA + bB ⇌ cC + dD
• the concentration-based equilibrium constant is:
• Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
• The product concentrations go in the numerator and the reactant concentrations go in the denominator.
• The coefficients from the balanced equation become exponents in the equilibrium expression. This is the current AP/Khan equilibrium expression form.

Consider:

• N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• The equilibrium expression is:
• Kc = [NH₃]² / ([N₂][H₂]³)
• Notice that:
• coefficient 2 → exponent 2
• coefficient 3 → exponent 3.

For gaseous equilibria, equilibrium can also be expressed using partial pressures.

• This constant is called Kp.

For:

• aA(g) + bB(g) ⇌ cC(g) + dD(g)
• the expression is:
• Kp = (PC)ᶜ(PD)ᵈ / (PA)ᵃ(PB)ᵇ
• where P represents the partial pressure of each gas.
• So the main distinction is:
• Kc → molar concentrations
• Kp → gas partial pressures

Pure solids and pure liquids do not appear in equilibrium expressions.

• Consider:
• CaCO₃(s) ⇌ CaO(s) + CO₂(g)
• The equilibrium expression is:
• K = [CO₂]
• or, for pressure:
• Kp = PCO₂
• CaCO₃(s) and CaO(s) are omitted.
• This happens because the effective concentrations, more precisely activities, of pure solids and liquids are constant under a given set of conditions.
• For AP Chemistry, simply remember:
• include gases and aqueous species
• omit pure solids and pure liquids

Consider:

• H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
• The pure liquid water is omitted.
• So the expression contains:
• K = [H⁺][OH⁻]
• This idea becomes extremely important in Unit 8 acid-base equilibrium.

The reaction quotient, represented by Q, has the exact same mathematical form as the equilibrium expression.

For:

• aA + bB ⇌ cC + dD
• Q = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
• The difference is when the values are measured.
• K uses concentrations or pressures at equilibrium.
• Q can be calculated using concentrations or pressures at any moment.
• Therefore:
• K tells you where equilibrium lies
• while:
• Q tells you where the system currently is

Comparing Q and K tells you which direction the reaction must proceed to reach equilibrium.

There are too few products relative to the equilibrium condition.

• The system proceeds toward:
• products
• or:
• to the right
• until Q increases to K.

There are too many products relative to equilibrium.

• The reaction proceeds toward:
• reactants
• or:
• to the left
• until Q decreases to K.

The system is already:

• at equilibrium
• and there is no net change.

Think:

• Q wants to become K.

If Q is too small:

• Q < K → move right → make more products → Q rises

If Q is too large:

• Q > K → move left → make more reactants → Q falls

Suppose:

• A ⇌ B
• and:
• Kc = 4.0
• At one moment:
• [A] = 0.50 M
• [B] = 1.00 M
• Then:
• Q = [B]/[A]
• Q = 1.00/0.50
• Q = 2.0
• Since:
• Q < K
• the reaction must proceed toward products.
• As B increases and A decreases, Q becomes larger until:
• Q = 4.0

If the equation changes, the equilibrium expression changes.

• For example:
• N₂O₄ ⇌ 2NO₂
• gives:
• K = [NO₂]²/[N₂O₄]
• But if we reverse the equation:
• 2NO₂ ⇌ N₂O₄
• the expression becomes:
• K = [N₂O₄]/[NO₂]²
• which is the reciprocal.
• This becomes even more important when studying properties of K.

Do not include pure solids or pure liquids in the equilibrium expression.

Do not forget that coefficients become exponents.

Do not confuse Q and K. They use the same form, but Q describes the current mixture while K describes equilibrium at a particular temperature.

Do not say Q<K means equilibrium shifts left. It shifts right.

Q < K → right

• Q > K → left
• Q = K → equilibrium

If the equilibrium concentrations or partial pressures are known, the equilibrium constant can be calculated by substituting those values into the appropriate equilibrium expression.

• Khan Academy currently includes calculations using both equilibrium concentrations for Kc and equilibrium partial pressures for Kp.

Consider:

• H₂(g) + I₂(g) ⇌ 2HI(g)
• Suppose at equilibrium:
• [H₂] = 0.20 M
• [I₂] = 0.20 M
• [HI] = 1.20 M
• Write the expression:
• Kc = [HI]² / ([H₂][I₂])
• Substitute:
• Kc = (1.20)² / [(0.20)(0.20)]
• Kc = 1.44 / 0.040
• Kc = 36
• A relatively large K indicates that equilibrium contains proportionally more products than reactants.

Consider:

• N₂O₄(g) ⇌ 2NO₂(g)
• Suppose equilibrium partial pressures are:
• PN₂O₄ = 0.50 atm
• PNO₂ = 1.00 atm
• Then:
• Kp = (PNO₂)² / PN₂O₄
• Kp = (1.00)² / 0.50
• Kp = 2.0

Sometimes the problem does not directly give all equilibrium concentrations.

• You may need to use stoichiometry first.
• Suppose:
• A ⇌ 2B
• Initially:
• [A] = 1.00 M
• [B] = 0

At equilibrium:

• [A] = 0.60 M
• A decreased by:
• 1.00 − 0.60 = 0.40 M
• According to the equation:
• 1 A → 2 B
• so B increases by:
• 2(0.40) = 0.80 M
• Therefore:
• [B]eq = 0.80 M
• Now:
• Kc = [B]²/[A]
• Kc = (0.80)²/0.60
• Kc ≈ 1.07
• This is why equilibrium calculations are often partly stoichiometry problems.

If a problem gives moles instead of concentration, remember:

• M = mol/L
• If the container volume is 1.00 L, then the numerical mole amount and molarity happen to be equal.
• But this is not true for every container.
• Suppose:
• 2.00 mol A are present in a 4.00 L vessel.
• Then:
• [A] = 2.00/4.00
• [A] = 0.500 M

Do not substitute raw mole values into Kc unless the situation legitimately allows the concentration to be determined that way.

For gas equilibrium, you may need Unit 3's Dalton's law:

• Ptotal = P1 + P2 + P3 + ...
• For example, if total equilibrium pressure and some individual partial pressures are known, you may find a missing partial pressure before calculating Kp.
• Khan Academy's current Unit 7 includes this type of equilibrium-pressure problem.

In rigorous thermodynamics, equilibrium constants are treated as dimensionless because they are based on activities.

• In many introductory classroom calculations, you may see apparent units if concentrations or pressures are substituted directly. For AP Chemistry, follow the convention and format expected by your course/reference sheet rather than trying to attach ordinary concentration units to K.
• The important thing is understanding what K means and how the expression is constructed.

Make sure every number you substitute represents an equilibrium value when calculating K.

Do not forget coefficients as exponents.

Do not substitute total pressure where a particular partial pressure is required.

Do not confuse Kc with Kp.

To calculate K:

• write expression first → determine equilibrium values → substitute → calculate

The magnitude of K tells us whether products or reactants are favored at equilibrium.

• It does not directly tell us how quickly equilibrium is reached. Reaction speed is a kinetics question.

If:

• K >> 1
• the numerator of the equilibrium expression is relatively large compared with the denominator.
• Therefore equilibrium contains proportionally more:
• products
• The reaction is said to be product-favored.
• For example:
• K = 4 × 10⁸
• indicates that equilibrium lies strongly toward products.
• However, this does not necessarily mean absolutely every reactant molecule disappears.

If:

• K << 1
• the denominator is relatively large.
• Equilibrium contains proportionally more:
• reactants
• The reaction is reactant-favored.
• For example:
• K = 2 × 10⁻⁹
• indicates equilibrium lies strongly toward reactants.

If K is roughly around 1, neither side overwhelmingly dominates.

• Significant amounts of both reactants and products may be present.

Suppose:

• K = 10¹⁰
• This tells you equilibrium strongly favors products.
• It does not mean the reaction occurs quickly.
• A reaction could have a very favorable equilibrium but still occur extremely slowly because of a large activation-energy barrier.
• This connects Unit 7 back to Unit 5.
• Kinetics → how fast
• Equilibrium → relative amounts once equilibrium is established

If:

• A ⇌ B
• has equilibrium constant:
• K
• then the reverse:
• B ⇌ A
• has:
• Kreverse = 1/Kforward
• Example:
• If:
• Kforward = 20
• then:
• Kreverse = 1/20 = 0.050
• This makes sense because a product-favored forward reaction becomes reactant-favored when viewed in reverse.

Suppose:

• A ⇌ B
• has:
• K = 4
• If the entire reaction is doubled:
• 2A ⇌ 2B
• the new equilibrium expression squares the old expression.
• Therefore:
• Knew = K²
• Knew = 4² = 16
• More generally, if a reaction is multiplied by a factor n:
• Knew = Kⁿ

If every coefficient is divided by 2:

• Knew = √K
• More generally:
• Knew = K^(1/n)
• when the equation is divided by n.

If two reactions are added together, their equilibrium constants are multiplied.

• Suppose:
• Reaction 1:
• A ⇌ B
• K₁ = 4
• Reaction 2:
• B ⇌ C
• K₂ = 3
• Add them:
• A ⇌ C
• B cancels.
• The overall equilibrium constant is:
• Koverall = K₁K₂
• Koverall = (4)(3)
• Koverall = 12
• These properties are part of Khan Academy's current equilibrium-constant lesson.

For a particular reaction, K has a specific value at a particular temperature.

• Changing concentrations, pressures, or volume may cause the system to shift, but these changes do not change K as long as temperature stays constant.

Changing temperature can change K.

• This distinction is extremely important.
• For example:
• adding reactant → equilibrium shifts
• but:
• K stays the same
• if temperature is unchanged.
• Heating or cooling → can change the equilibrium constant.

Do not interpret large K as fast reaction.

Do not simply change K when concentration changes.

If you reverse the equation, take the reciprocal of K.

If you multiply coefficients by n, raise K to the nth power.

If you add reactions, multiply their K values.

Large K → products favored

• Small K → reactants favored
• And for reaction manipulation:
• reverse → 1/K
• multiply equation by n → Kⁿ
• add equations → multiply K values

For:

• aA + bB ⇌ cC + dD
• write:
• products over reactants
• and:
• coefficients become exponents
• Include mainly:
• aqueous species and gases
• Omit:
• pure solids and pure liquids
• For example:
• CaCO₃(s) ⇌ CaO(s) + CO₂(g)
• gives:
• Kp = PCO₂
• not an expression containing the solids.

A very easy memory tool is:

• Q always wants to become K.